Misc 5 - If coordinates of point A, B, C, D be (1, 2, 3), (4, 5, 7)

Misc 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Misc 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

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Question 3 If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.Angle between a pair of lines having direction ratios 𝑎1, 𝑏1, c1 and 𝑎_2 , 𝑏_2, 𝑐_2 is given by cos θ = |(𝒂_𝟏 𝒂_𝟐 + 𝒃_𝟏 𝒃_𝟐 + 𝒄𝟏𝒄𝟐)/(√(〖𝒂_𝟏〗^𝟐 + 〖𝒃_𝟏〗^𝟐+〖 𝒄_𝟏〗^𝟐 ) √(〖𝒂_𝟐〗^𝟐 + 〖𝒃_𝟐〗^𝟐+〖 𝒄_𝟐〗^𝟐 ))| A line passing through A (𝑥_1, 𝑦_1, 𝑧_1) and B (𝑥_2, 𝑦_2, 𝑧_2) has direction ratios (𝑥_2 − 𝑥_1), (𝑦_2 − 𝑦_1), (𝑧_2 − 𝑧_1) AB A (1, 2, 3) , B (4, 5, 7) Direction ratios of AB (4 − 1), (5 − 2),(7 − 3) = 3, 3, 4 ∴ 𝒂1 = 3, 𝒃1 = 3, 𝒄1 = 4 CD C (−4, 3, −6) ,D (2, 9, 2) Direction ratios of CD (2 − (–4)), (9 − 3),(2 – (–6)) = 6, 6, 8 ∴ 𝒂2 = 6, 𝒃2 = 6, 𝒄2 = 8 Now, cos θ = |(𝑎_1 𝑎_2 + 𝑏_1 𝑏_2 + 𝑐1𝑐2)/(√(〖𝑎_1〗^2 + 〖𝑏_1〗^2+〖 𝑐_1〗^2 ) √(〖𝑎_2〗^2 + 〖𝑏_2〗^2+〖 𝑐_2〗^2 ))| cos θ = |(3 × 6 + 3 × 6 + 4 × 8 )/(√(32 + 32 + 42) √(62 + 62 + 82))| = |(18 + 18 + 32 )/(√(9 + 9 + 16) √(36 + 36 + 64))| = |68/(√34 √136)| = |68/(√34 √(4 × 34))| = |68/(√34 × √4 × √34)| = |68/(√34 × √34× √4)| = |68/(34 × 2 )| = |68/68| = 1 ∴ cos θ = 1 So, θ = 0° Therefore, angle between AB and CD is 0° .

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo